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Last updated date: 02nd Dec 2023
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MVSAT Dec 2023

Choose the correct statement to correct the following (if required), for \[2{{x}^{2}}+bx+8\] to have non-real roots.
(A) \[-8 < b < 8\]
(B) \[-4 < b < 8\]
(C) \[0 < b < 8\]
(D) \[8 < b < 0\]

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Hint: We are given a quadratic equation with a variable ‘b’ and that the given quadratic equation has non – real roots and we are asked to find the value of ‘b’ such that the given condition is met. And we have to select the most appropriate option from the given. The given quadratic equation is in the form, \[a{{x}^{2}}+bx+c\]. We know that, for non-real roots we have the discriminant as, \[D={{b}^{2}}-4ac < 0\]. We will then substitute the corresponding values and we will then check and find the range in which the value of ‘b’ lies such that the given quadratic equation gives non-real roots.

Complete step-by-step solution:
According to the given question, we are given a quadratic equation which has non-real roots. And we are asked in the question, to find the value of ‘b’ such that the conditions are met too.
The quadratic equation that we have is,
We can see that the given quadratic equation is of the form, \[a{{x}^{2}}+bx+c\].
We know that for non – real roots, we have the discriminant given by,
\[D={{b}^{2}}-4ac < 0\]
We will now substitute the values in the above formula of the discriminant and we get,
\[\Rightarrow {{b}^{2}}-4\times 2\times 8 < 0\]
\[\Rightarrow {{b}^{2}}-64 < 0\]
Now, applying the formula \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\], we get,
\[\Rightarrow \left( b-8 \right)\left( b+8 \right) < 0\]
We have the interval points as, -8 and 8. We will now check for the increasing and decreasing in these intervals and so we have,
Intervals \[\left( -\infty ,-8 \right)\]\[\left( -8,8 \right)\]\[\left( 8,\infty \right)\]
Value of function+ve-ve+ve

So, for non-real roots, the value of \[b\in \left( -8,8 \right)\].
Therefore, the correct answer is (A) \[-8 < b < 8\].

Note: The quadratic equation with non – real roots have negative value to the square and then a new term is introduced, which is, iota which is complex numbers. Also, the formula for discriminant should be correct and while substituting the values in the formula make sure that the values don’t get interchanged, else the entire solution will get wrong.